Cable volt drop9
All cables have resistance, and when current flows in them
this results in a volt drop. Hence, the voltage at the load is lower than the
supply voltage by the amount of this volt drop.
The volt drop may be calculated using the basic Ohm's law
formula
| |
U = I x R
| |
| where | U is the cable volt drop (V |
| I is the circuit
current (A), and | |
| R is the circuit resistance W(Ohms) | |
Unfortunately, this simple formula is seldom of
use in this case, because the cable resistance under load conditions is not easy
to calculate.
[525-01-03] indicates that the voltage at any load must never
fall so low as to impair the safe working of that load, or fall below the level
indicated by the relevant British Standard where one applies.
[525-01-02] indicates that these requirements will he met if
the voltage drop does not exceed 4% of the declared supply voltage. If the
supply is single-phase at the usual level of 240 V, this means a maximum volt
drop of 4% of 240 V which is 9.6 V, giving (in simple terms) a load voltage as
low as 230.4 V. For a 415 V three-phase system, allowable volt drop will be 16.6
V with a line load voltage as low as 398.4 V.
It should be borne in
mind that European Agreement RD 472 S2 allows the declared supply voltage of 230
V to vary by +10% or -6%. Assuming that the supply voltage of 240 V is 6% low,
and allowing a 4% volt drop, this gives permissible load voltages of 216.6 V for
a single-phase supply, or 374.5 V (line) for a 415 V three-phase
supply.
To calculate the volt drop for a particular cable we use {Tables 4.6, 4.7 and 4.9}. Each current rating table
has an associated volt drop column or table. For example, multicore sheathed
non-armoured P.V.C. insulated cables are covered by {Table
4.7} for current ratings, and volt drops. The exception in the
Regulations to this layout is for mineral insulated cables where there are
separate volt drop tables for single- and three-phase operation, which are
combined here as {Table
4.9}.
Each cable rating in the Tables of [Appendix 4] has a
corresponding volt drop figure in millivolts per ampere per metre of
run (mV/A/m). Strictly this should be mV/(A m), but here we shall
follow the pattern adopted by BS 7671: 1992. To calculate the cable volt
drop:
1. - take the
value from the volt drop table (mV/A/m)
For example, if a 4 mm² p.v.c. sheathed circuit feeds a 6 kW
shower and has a length of run of 16 m, we can find the volt drop
thus:
From {Table 4.7}, the
volt drop figure for 4 mm² two-core cable is 11
mV/A/m.
Cable current is calculated from
|
I =
|
P = 6000 A
|
|
U 240-----
|
Volt drop is then
|
11 x 25 x 16 V
|
|
1000
|
Since the permissible volt drop is 4% of 240 V, which is 9.6
V, the cable in question meets volt drop requirements. The following examples
will make the method clear.
Example
4.5
The table concerned here is {4.7}, which shows a figure of 7.3
mV/A/m for 6 mm² twin with protective conductor pvc insulated and sheathed
cable. The actual circuit current is 12.5 A, and the length of run is 14
m.
Volt drop =
|
7.3 x 12.5 x 14
V
|
= 1.28 V
|
1000
|
| Maximum permissible volt drop is |
4% of 240 V =
|
4 of 240 V
|
|
100
|
If a 14 m run gives a volt drop of 1.28 V,
the length of run for a 9.6 V drop will be: |
9.6 x 14m
|
|
1.28
|
Example
4.6
The Table concerned here is {4.7} which shows a volt drop figure
for 4.0 mm² cable of 11mV/A/m, with the current and the length of run remaining
at 12.5 A. and 14 m respectively.
Volt drop =
|
11 x 12.5 x 14 V
|
= 1.93 V
|
1000
|
| Maximum permissible volt drop is |
4% of 240 V =
|
4 of 240 V
|
|
100
|
If a 14 m run gives a volt drop of 1.93 V,
the length of run for a 9.6 V drop will be: |
9.6 x 14m
|
|
1.93
|
Example 4.7
In neither case is there any change in cable sizes, the
selected cables being 6 mm² in the first case and 4 mm² in the second. Solutions
are thus the same as those in {Examples 4.5 and
4.6} respectively.
Example 4.8
This time we have a mineral insulated p.v.c. sheathed cable,
so volt drop figures will come from {Table
4.9}. This shows 9.1 mV/A/m for the 4 mm² cable selected,
which must be used with the circuit current of 15.3 A and the length of run
which is 20 m.
Volt drop =
|
9.1 x 15.3 x 20 V
|
= 2.78 V
|
1000
|
| Maximum permissible volt drop is |
4% of 415 V =
|
4 of 415V
|
|
100
|
Maximum length of run for this circuit
with the same cable size and type will be: |
16.6 x 20m
|
|
2.78
|
The 'length of run' calculations carried out in these
examples are often useful to the electrician when installing equipment at
greater distances from the mains position.
It is important to appreciate that the allowable volt drop of
4% of the supply voltage applies to the whole of an installation. If an
installation has mains, sub-mains and final circuits, for instance, the volt
drop in each must be calculated and added to give the total volt drop as
indicated in {Fig 4.10}.
All of our work in this sub-section so far has assumed that
cable resistance is the only factor responsible for volt drop. In fact, larger
cables have significant self inductance as well as resistance. As we shall see
in Chapter
5 there is also an effect called impedance which is made up of
resistance and inductive reactance (see {Fig
5.8(a)}).
Inductive reactance
|
XL
|
2(pi)fL
| |
| where |
XL
|
= |
inductive reactance in Wohms
|
| (pi) | = |
the mathematical constant 3.142
| |
| f | = |
the system frequency in hertz (Hz)
| |
| L | = |
circuit self inductance in henrys
(H)
|
 |
Fig 4.10 Total volt drop in large
installations
For small cables, the self inductance is such that the
inductive reactance, is small compared with the resistance. Only with cables of
cross-sectional area 25 mm² and greater need reactance be considered. Since
cables as large as this are seldom used on work which has not been designed by a
qualified engineer, the subject of reactive volt drop component will not be
further considered here.
If the actual current
carried by the cable (the design current) is less than the rated value, the
cable will not become as warm as the calculations used to produce the volt drop
tables have assumed, The Regulations include (in [Appendix 4]) a very
complicated formula to be applied to cables of cross-sectional area 16 mm² and
less which may show that the actual volt drop is less than that obtained from
the tables. This possibility is again seldom of interest to the electrician, and
is not considered here.
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